Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

<image>

You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
class Segment {
 public:
  double left, right;
  Segment(double l, double r) : left(l), right(r) {}
  Segment() {}
  double length() const { return (right - left); }
  bool include(double x) const { return (x >= left && x <= right); }
};
int binary_search_a(int left, int right, double x);
int binary_search_b(int left, int right, double x);
const int INF = 1e9 + 7;
const int MAX = 2e5 + 5;
Segment segments[MAX];
double sum[MAX];
int main() {
  int sy, a, b;
  int n, q;
  while (scanf("%d %d %d", &sy, &a, &b) != EOF) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%lf %lf", &segments[i].left, &segments[i].right);
    }
    segments[0] = Segment(0, 0);
    segments[n + 1] = Segment(INF, INF);
    scanf("%d", &q);
    sum[0] = 0;
    for (int i = 1; i <= n; i++) {
      sum[i] = sum[i - 1] + segments[i].length();
    }
    double Px, Py;
    while (q--) {
      scanf("%lf %lf", &Px, &Py);
      double k = (Py - sy) / Py;
      double A0 = Px - (Px - a) / k;
      double B0 = Px - (Px - b) / k;
      int ia = binary_search_a(1, n, A0);
      int ib = binary_search_b(1, n, B0);
      double res = 0, sumc = 0, suml = 0, sumr = 0;
      if (false && ia == ib && segments[ia + 1].include(A0) &&
          segments[ib + 1].include(B0)) {
        res = b - a;
      } else {
        sumc = sum[ib - 1] - sum[ia - 1];
        suml = (ia >= 1 ? max(segments[ia - 1].right - A0, 0.0) : 0);
        sumr = (ib <= n ? max(B0 - segments[ib].left, 0.0) : 0);
        res = (sumc + suml + sumr) * k;
      }
      printf("%.6f\n", res);
    }
  }
  return 0;
}
int binary_search_a(int left, int right, double x) {
  while (left <= right) {
    int mid = (left + right) / 2;
    if (x >= segments[mid].left) {
      left = mid + 1;
    } else {
      right = mid - 1;
    }
  }
  return left;
}
int binary_search_b(int left, int right, double x) {
  while (left <= right) {
    int mid = (left + right) / 2;
    if (x > segments[mid].right) {
      left = mid + 1;
    } else {
      right = mid - 1;
    }
  }
  return left;
}